Question

The relation \(R = \{ (a,b): b = a + 2 \}\) defined in the set \(A = \{1, 2, 3, 4, 5\}\) is

• A. not reflexive and symmetric, but transitive
• B. not reflexive and transitive, but symmetric
• C. not symmetric and transitive, but reflexive
• D. not reflexive, not symmetric and not also transitive

Hint:

When testing properties of a relation on a finite set, it is often easiest to first list all the elements of the relation. Then, check for reflexivity (all (a,a) pairs), symmetry (if (a,b) exists, does (b,a) exist?), and transitivity (if (a,b) and (b,c) exist, does (a,c) exist?) by examining the list.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To determine the properties of the relation R on the set A, we need to check for three conditions: reflexivity, symmetry, and transitivity.
First, let's list the ordered pairs in the relation R based on the rule \(b = a + 2\).
For \(a=1, b=1+2=3\), so \((1,3) \in R\).
For \(a=2, b=2+2=4\), so \((2,4) \in R\).
For \(a=3, b=3+2=5\), so \((3,5) \in R\).
For \(a=4, b=4+2=6\), which is not in set A.
For \(a=5, b=5+2=7\), which is not in set A.
So, the relation is \(R = \{ (1,3), (2,4), (3,5) \}\).
Step 2: Key Formula or Approach:
Reflexivity: A relation is reflexive if \((a,a) \in R\) for every \(a \in A\).
Symmetry: A relation is symmetric if \((a,b) \in R\) implies \((b,a) \in R\) for all \(a,b \in A\).
Transitivity: A relation is transitive if \((a,b) \in R\) and \((b,c) \in R\) implies \((a,c) \in R\) for all \(a,b,c \in A\).
Step 3: Detailed Explanation:
Checking for Reflexivity:
For R to be reflexive, \((1,1), (2,2), (3,3), (4,4), (5,5)\) must be in R.
However, none of these pairs are in R. For example, for \(a=1\), \(b=1 \neq 1+2\), so \((1,1) \notin R\).
Thus, R is not reflexive.
Checking for Symmetry:
For R to be symmetric, if \((a,b) \in R\), then \((b,a)\) must also be in R.
We have \((1,3) \in R\). For symmetry, \((3,1)\) should be in R.
But for \((3,1)\), we have \(a=3, b=1\). The condition \(b = a+2\) becomes \(1 = 3+2\), which is false. So \((3,1) \notin R\).
Thus, R is not symmetric.
Checking for Transitivity:
For R to be transitive, if \((a,b) \in R\) and \((b,c) \in R\), then \((a,c)\) must be in R.
Let's check for pairs that satisfy the condition. We have \((1,3) \in R\). We also have a pair starting with 3, which is \((3,5) \in R\).
So we have \((a,b)=(1,3)\) and \((b,c)=(3,5)\). For transitivity, \((a,c)=(1,5)\) must be in R.
Let's check if \((1,5)\) satisfies the condition \(b=a+2\). Here \(a=1, b=5\). The condition is \(5 = 1+2\), which is false. So \((1,5) \notin R\).
Thus, R is not transitive.
Step 4: Final Answer:
The relation R is not reflexive, not symmetric, and not transitive.